Friction

Question 1:

Question 2:

It is not necessary that the friction coefficient is always less than 1. When the friction is stronger than the normal reaction force, the coefficient of friction is greater than 1. For example, silicon rubber has the coefficient of friction greater than 1.

Question 3:

It is easier to push a heavy block from behind than from the top because when we try to push a heavy block from the top, we increase the normal reaction force, which, in turn, increases the friction between the object and the ground (see the figure).

Question 4:

The person started with zero initial velocity, covered a 1 km distance and ended with zero velocity, so the acceleration is zero. Hence, the average friction force is zero.

Question 5:

It is difficult to walk on solid ice because the coefficient of friction between our foot and ice is very less; hence, a person trying to walk on solid ice may slip.

Question 6:

No, we cannot accelerate or stop a car on a frictionless horizontal road. The car will not move on a frictionless surface because rolling is not possible without friction.

Question 7:

In the first case, the normal reaction force is equal to the weight of the stone, hence the stone slides easily because the friction force is very less. However, in the second case, a small piece of wood is sandwiched, which increases the normal reaction force on the wood due to the weight of the door. Hence, greater the normal reaction force on the wood, the greater will be the frictional force between wood and the floor.

Question 8:

(a) The coefficient of friction between the card and the coin should be small.
(b) The coin should be heavy.
(c) If the card is pushed gently, the experiment fails because the frictional force gets more to time to act and it may gain some velocity and move with the card.

Question 9:

No, a tug of war cannot be won on a frictionless surface because the tension in the rope on both the sides of both the teams will be same. So, to win, one of the teams must apply some greater force, which is the force of friction. 

Question 10:

The normal reaction force on a level road is mg, whereas on an inclined plane it is mg cos θ, which means that on an incline road the friction force between the tyre and the road is less. Hence, tyres have less grip on an incline plane and better grip on a level road.

Question 11:

By throwing the bag in one direction, we gain some velocity in the opposite direction as per the law of conservation of linear momentum. In this way we can come out of the ice easily.

Question 1:

The coefficient of friction increases between two highly smooth surfaces because the atoms of both the materials come very closer to each and the number of bonds between them increase.

Question 2:

(b) decrease

According to the first law of limiting friction,
f = μN
where f  is the frictional force
         N is the normal reaction force
         μ is the coefficient of static friction

and

N = mg ï¼ Fcosθ
where m is the mass of the body
         F is the contact force acting on the body

If we decrease the angle between this contact force and the vertical, then Fcosθ increases and the normal reaction force (N) as well as the frictional force (f) decrease.

Question 3:

(b) smaller friction

According to the first law of the limiting friction,
f = μN
where f is the frictional force
         μ is the coefficient of friction
         N is the normal reaction force

When we take smaller steps on ice, the normal reaction force exerted by the ice is small. Therefore, the smaller steps ensure smaller friction.

Question 4:

(c) Mg ≤ F ≤ Mg $\sqrt{1+{\mathrm{\mu }}^2}$

Let T be the force applied on an object of mass M.

If T = 0, F_min = Mg.

If T is acting in the horizontal direction, then the body is not moving.
∴ $T=\mathrm{\mu }\left(mg\right)$

$$\begin{gathered} F_{\max } =\sqrt{(Mg{)}^2+(T{)}^2} \\ =\sqrt{(Mg{)}^2+(\mathrm{\mu }Mg{)}^2} \end{gathered}$$

Thus, we have:
$Mg\le F\le Mg\sqrt{1+(\mathrm{\mu }{)}^2}$

Question 5:

d) >500 N for time ∆t₁ and ∆t₃ and 500 N for ∆t_2.

During the time interval ∆t₂, the scooter is moving with a constant velocity, which implies that the force exerted by the seat on the man is 500 N (for balancing the weight of the man).

During the time interval ∆t₁ and ∆t₃,the scooter is moving with constant acceleration and deceleration, which implies that a frictional force is also applied. Therefore, the net force exerted by the seat on the man should be >500 N.

Question 6:

(d) the system cannot remain in equilibrium.


Since the wall is smooth and the surface of A and B in contact are rough, the net vertical force on the system is in the downward direction. Hence, the system cannot remain in equilibrium.

Question 7:

 (a) is upward


The normal reaction force on the system (comprising of wall and contact surface of A and B) is provided by F. As can be seen from the figure, the weight of A and B is in the downward direction. Therefore, the frictional force f_A and f_BA (friction on B due to A) is in upward direction.

Question 8:

(c) is same for both cars

Given: both the cars have same initial speed.
Let the masses of the two cars be m₁ and m_2.

Frictional force on car with mass m₁ = μm₁ g
So, the deceleration due to frictional force = $\frac{\mathrm{\mu }{m}_1\mathrm{g}}{m_1}=\mathrm{\mu g}$

Frictional force on car with mass m₂ = μm₂ g
So, the deceleration due to frictional force = $\frac{\mathrm{\mu }{m}_2\mathrm{g}}{m_2}=\mathrm{\mu g}$
As both the acceleration are same, from the second equation of motion
$s=ut+\frac{1}{2}a{t}^2$

Thus, we can say that both the cars have same minimum stopping distance.

Question 9:

(b) apply the brakes hard enough to just prevent slipping

When we apply hard brakes just enough to prevent slipping on wheels, it provides optimum normal reaction force, which gives the maximum friction force between tyres of the car and the road.

Question 10:

(d) all of three are possible.

      
We know that
N = mg cos θËš
f_max = μN = μmg cos θ
where N  =  normal reaction force
          f_max = frictional force
         θ =  angle of inclination
         μ = coefficient of friction

When the block just begins to slide, it means
mg sin θ = f_max
mg sin θ ₌ μmg cos θ
          μ = tan θ
and the coefficient of friction depends on the angle of inclination (θ) and does not depend on mass.

Now consider the block sliding condition:
mg sin θ − f_max = ma
mg sin θ−μmg cos θ = ma
∴ a = g(sin θ − μ cos θ)

From the above equation it is clear that acceleration does not depend on the mass but depends on θ and μ.

Question 1:

(a) μ < μ', M < M'
 
Let T be the force applied by the boy on the block. 


 


Free body diagram for the box:

The condition for preventing the slide is
     f_max > T

μ'M'g > T         (i)

Now see the free body diagram of a boy of mass M:

     
f_max = μmg

The condition for preventing the slide is
f_max > T
μmg > T 

The condition for sliding the entire system (block and boy) is
f' > f (block is not slide)
μ'M'g > μmg
μ'M' > μm
μ < μ'
m < M'

Question 2:

(a) F > F_N
(b) F > f
(d) F_N − f < F < F_N + f

The system is in equilibrium condition when F = f.
Hence, the net horizontal force is zero.

f = μF_N
F > F_N
f = F_Nand 0 ≤ μ ≤ 1
Therefore, we can say that F > f. So the net horizontal force is nonzero.

F > f, and so the net horizontal force is zero.

F_N > f$\Rightarrow$F_N > μF_N$\Rightarrow$μ < 1

Here, the given relation between F and f i.e 

F > f and f = μF_N  will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.

F_N − f < F < F_N + f
∵ f = μF_N
$$\begin{gathered} \frac{f}{\mu }-f<F<\frac{f}{\mu }+f \\ f\left(\frac{1-\mu }{\mu }\right)<F<f\left(\frac{1+\mu }{\mu }\right) \end{gathered}$$
For the above relation, we can say that F ≠ f and so the net horizontal force is nonzero.

Question 3:

(b) the force of friction between the bodies is zero
(d) the bodies may be rough but they don't slip on each other

The contact force exerted by a body A on another body B is equal to the normal force between the bodies. Therefore, we can conclude that the force of friction between the bodies is zero or the bodies may be rough but they don't slip on each other.

Question 4:

(b) Coefficient of static friction is always greater than the coefficient of kinetic friction.
(c) Limiting friction is always greater than the kinetic friction.
(d) Limiting friction is never less than the static friction.

All the above statements are correct. The static friction is sometimes less than the kinetic friction.

Question 5:

(c) The graph is a straight line of slope 45° for small F and a straight line parallel to the F-axis for large F.
(d) There is a small kink on the graph.



When force F is applied on the block, the force of friction f comes into play. As we increase the  applied F,  the static friction force adjusts itself to become  (equal) to the applied force F  and goes upto  its maximum value equal to limiting friction force.After this ,it is treated as a constant force (i.e . now its value does not change until and unless the body starts moving). If the applied force F is greater than the limiting friction force, then the kinetic friction force comes into play at that time. The kinetic friction force is always less than the limiting friction force.

Question 1:

(a) is towards east if the vehicle is accelerating
(b) is zero if the vehicle is moving with a uniform velocity

When the vehicle is accelerating, the force is applied (by the tyre on the road) in west direction .That causes a net resultant  frictional force acting in east direction. Due to this  force of friction only ,the  car is moving in east direction.

When the vehicle is moving with a uniform velocity, the force of friction on the wheels of the vehicle by the road is zero.

Question 2:

Let m be the mass of the body.



From the free body diagram,
R − mg = 0
(where R is the normal reaction force and g is the acceleration due to gravity)
⇒ R = mg                 (1)
Again ma − μ_kR = 0
(where μ_k is the coefficient of kinetic friction and a is deceleration)
or ma = μ_kR
From Equation (1),
ma =  μ_kmg
⇒ a = μ_kg
⇒ 4 = μ_kg
$\Rightarrow {\mu }_{\mathrm{k}}=\frac{4}{g}=\frac{4}{10}=0.4$

Hence, the coefficient of the kinetic friction between the block and the plane is 0.4.

Question 3:

Friction force acting on the block will decelerate it.
Let the deceleration be 'a'.

Using free body diagram


R − mg = 0
(where R is the normal reaction force)
⇒ R = mg                 (1)
Again, ma − μ_kR = 0
(where μ_k is the coefficient of kinetic friction)
From Equation (1),
⇒ ma = μ_kmg
⇒ a = μ_kg = 0.1 × 10
      = 1 m/s²
Given:
initial velocity, u = 10 m/s
final velocity, v = 0 m/s (block comes to rest)
                       a = −1 m/s² (deceleration)
Using equation of motion v² $-$u² = 2as
(where s is the distance travelled before coming to rest)
$s=\frac{v^2-u^2}{2a}$
On substituting the respective values, we get
$$\begin{gathered} =0-\frac{{10}^2}{2} \left(-1\right) \\ =\frac{100}{2}=50 \mathrm{m} \end{gathered}$$
Therefore, the block will travel 50 m before coming to rest.

Question 4:

A block of mass m is kept on a horizontal table. If force is applied on the block, a friction force will be there: p → frictional force and F → applied force


So, friction force is equal to the applied force. One of the case is that the friction force is equals to zero when the applied force is equal to zero.

Question 5:

Free body diagram for the block is as follows:


From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ                     (1)

For the block, u = 0 m/s, s = 8 m and t = 2 s.
According to the equation of motion
$s=ut+\frac{1}{2}a{t}^2$
$$\begin{gathered} s=0+\frac{1}{2}a{2}^2 \\ 8=2a \end{gathered}$$
a = 4 m/s²

Again,
μ_kR + ma − mg sin θ = 0
(where μ_k is the coefficient of kinetic friction)
From Equation (1):
μ_kmg cos θ + ma − mg sin θ = 0
⇒ m (μ_kg cos θ + a − g sin θ) = 0
⇒ μ_k × 10 × cos 30° = g sin 30° − a
$$\begin{gathered} \Rightarrow {\mathrm{\mu }}_{\mathrm{k}}\times 10\frac{\sqrt{3}}{2}=10\times \left(\frac{1}{2}\right)-4 \\ \Rightarrow \left(5\sqrt{3}\right){\mathrm{\mu }}_{\mathrm{k}}=1 \\ \Rightarrow {\mathrm{\mu }}_{\mathrm{k}}=\left(\frac{1}{5\sqrt{3}}\right)=0.11 \end{gathered}$$

Therefore, the coefficient of kinetic friction between the block and the surface is 0.11.

Question 6:

Free body diagram of the block for this case is as follows:



From the adove diagram:
F − ma − μ_kR + mg sin 30° = 0
4 − 4a − μ_kR + 4g sin 30° = 0               (1)
R − 4g cos 30° = 0                               (2)
⇒ R = 4g cos 30° = 0

Substituting the values of R in Equation (1) we get
4 − 4a − 0.11 × 4g cos 30° + 4g sin 30° = 0
$\Rightarrow 4-4a-0.11\times 4\times 10\times \frac{\sqrt{3}}{2}+4\times 10\times \frac{1}{2}=0$
4 − 4a − 3.81 + 20 = 0
4 − 4a − 3.18 + 20 = 0
a ≈ 5 m/s²

For the block, u = 0, t = 2 s and a = 5 m/s².
According to the equation of motion,
              â€‹    â€‹$s=ut+\frac{1}{2}a{t}^2$
                  $$\begin{gathered} =0+\left(\frac{1}{2}\right)5\times 2^2 \\ =10 \mathrm{m} \end{gathered}$$

Therefore, the block will move 10 m.

Question 7:

(a) To make the block move up the incline, the applied force should be equal and opposite to the net force acting down the incline.
Applied force = μR + 2g sin 30°          (1)
(where μ is the coefficient of static friction)

R = mg cos 30°

Substituting the respective values in Equation (1), we get
$=0.2\times \left(9.8\right)\sqrt{3}+2\times 9.8\times \left(\frac{1}{2}\right)$
3.39 + 9.8 $\approx$ 13 N



With this minimum force, the body moves up the incline with a constant velocity as the net force on it is zero.

(b) Net force acting down the incline is given by
F = 2g sin 30° − μR


   $$\begin{gathered} =2\times 9.8\times \frac{1}{2}-3.99 \\ =6.41 \mathrm{N} \end{gathered}$$

Because F = 6.41 N, the body will move down the incline with acceleration, hence the force required is zero.

Question 8:

Using the free body diagram,



g = 10 m/s², m = 2 kg, θ = 30 and μ = 0.2

R − mg cos θ − F sin θ = 0
⇒ R = mg cos θ + F sin θ              (1)
and
mg sin θ + μR − F cos θ = 0
⇒ mg sin θ + μ(mg cos θ + F sin θ) − F cos θ = 0
⇒ mg sin θ + μmg cos θ + μF sin θ − F cos θ = 0
$$\begin{gathered} \Rightarrow F=\frac{\left(mg \sin \mathrm{\theta }+\mathrm{\mu } mg \cos \mathrm{\theta }\right)}{\left(\mathrm{\mu } \sin \mathrm{\theta }-\cos \mathrm{\theta }\right)} (\mathrm{\theta }=30°) \\ =\frac{\left(2\times 10\times \left({\displaystyle \frac{1}{2}}\right)+0.2\times 2\times 10\times {\displaystyle \frac{\sqrt{3}}{2}}\right)}{0.2\times \left({\displaystyle \frac{1}{2}}\right)-\left({\displaystyle \frac{\sqrt{3}}{2}}\right)} \\ =\frac{13.464}{0.76}=17.7 \mathrm{N}=17.5 \mathrm{N} \end{gathered}$$

Therefore, while pushing the block to move up on the incline, the required force is 17.5 N.

Question 9:

Let m be the mass of the boy.

            

From the above diagram:
R − mg cos 45° = 0
$R=mg \cos 45°=\frac{mg}{\sqrt{2}} \left(1\right)$
Net force acting on the boy, making him slide down
= mg sin 45° − μR
= mg sin 45° − μmg cos 45°
$$\begin{gathered} =m\times 10\times \left(\frac{1}{\sqrt{2}}\right)-0.6\times m\times 10\times \left(\frac{1}{\sqrt{2}}\right) \\ =m\left(5\sqrt{2}-3\sqrt{2}\right)=m\times 2\times \sqrt{2} \end{gathered}$$
The acceleration of the boy $=\frac{\mathrm{Force}}{\mathrm{Mass}}$
                     $$\begin{gathered} =\frac{m\left(2\sqrt{2}\right)}{m} \\ =2\sqrt{2} \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Question 10:

Let a be the acceleration of the body sliding down. 



From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ                      (1)
and
ma + mg sin θ − μR = 0
⇒ $a=\frac{mg(\sin \theta -\mathrm{\mu } cos\theta \mathit{)}}{m}=g(\sin \theta -\mathrm{\mu } cos\theta \mathit{)}$

For the first half metre, u = 0, s = 0.5 m and t = 0.5 s.

According to the equation of motion,
        v = u + at
           = 0 + (0.5)4 = 2 m/s
        $s=ut+\frac{1}{2}a{t}^2$
$$\begin{gathered} 0.5=0+\frac{1}{2}a{\left(0.5\right)}^2 \\ \Rightarrow a=4 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

For the next half metre, u = 2 m/s, a = 4 m/s² and s = 0.5.
$\Rightarrow 0.5=2t+\left(\frac{1}{2}\right)4t^2$
⇒ 2t² + 2t − 0.5 = 0
⇒ 4t² + 4t − 1 = 0
$$\begin{gathered} \Rightarrow t=\frac{-4\pm \sqrt{16+16}}{2\times 4} \\ =\frac{1.656}{8}=0.2027 \end{gathered}$$
Therefore, the time taken to cover the next half metre is 0.21 s.

Question 11:

Let
 f  be the applied force,
R  be the normal reaction force and
F  be the frictional force.



The coefficient of static friction is given by
$u=\tan \mathrm{\lambda }=\frac{\mathit{F}}{\mathit{R}}$
(where λ is the angle of friction)

When F = μR, F is the limiting friction (maximum friction). When applied force increases and the body still remains still static then the force of friction increases up to its maximum value equal to limiting friction (μR).

$$\begin{gathered} F\mathit{<}\mu R \\ \therefore \tan \mathrm{\lambda }=\frac{\mathit{F}}{\mathit{R}}\mathit{\le }\frac{\mathit{\mu }\mathit{R}}{\mathit{R}} \end{gathered}$$
⇒ tan λ ≤ μ
⇒ λ ≤ tan⁻¹ μ

Question 12:

         
From the above diagrams:
T + ma − mg = 0
T + 0.5a − 0.5 g = 0                       (1)
μR + ma + T₁ − T = 0
μR + 1a + T₁ − T = 0                      (2)
μR + 1a − T₁ = 0
μR + a = T₁                                    (3)

From Equations (2) and (3) we have
μR + a = T − T₁
⇒ T − T₁ = T₁
⇒ T = 2T₁

So, Equation (2) becomes
μR + a + T₁− 2T₁ = 0
⇒ μR + a − T₁ = 0
⇒ T₁= μR + a
        = 0.2g + a                       (4)

and Equation (1) becomes
2T₁ + 0.5a − 0.5g = 0

$$\begin{gathered} \Rightarrow T_1=\frac{0.5g-0.5a}{2} \\ =0.25g-0.25a \left(5\right) \end{gathered}$$

From Equations (4) and (5)
0.2g + a = 0.25g − 0.25a

$$\begin{gathered} \Rightarrow a=\frac{0.05}{1.25}\times 10 \\ =0.4\times 10 m/s^2 \left[g=10 m/s^2\right] \end{gathered}$$

Therefore,
(a) the acceleration of each 1 kg block is 0.4 m/s²,
(b) the tension in the string connecting the 1 kg blocks is
   T₁ = 0.2g + a + 0.4 = 2.4 N
​    and
(c) the tension in the string attached to the 0.5 kg block is 
    T = 0.5g − 0.5a
       = 0.5 × 10 − 0.5 × 0.4
       = 4.8 N.

Question 13:

From the free body diagram:
μ₁R + m₁a − F = 0
μ₁R + 1 − 16 = 0 (R = mg cos θ)



⇒ μ₁(2g) + (−15) = 0
${\mathrm{\mu }}_1=\frac{15}{20}=0.75$
Again, 
μ₂R₁ + ma = F − mg sin θ = 0
μ₂R₁ + 4 × 0.5 = 16 − 4g sin 30° = 0
R₁= mg cos θ      (θ = 30°)
$$\begin{gathered} \Rightarrow {\mathrm{\mu }}_2\left(20\sqrt{3}\right)+2+16-20=0 \\ \Rightarrow {\mathrm{\mu }}_2=\frac{2}{20}\sqrt{3}=\frac{1}{17.32} \\ =0.057=0.06 \end{gathered}$$

Therefore, the friction coefficients at the two contacts with blocks are
 μ₁ = 0.75 and μ₂ = 0.06.

Question 14:

Consider that a 15 kg object is moving downward with an acceleration a.

From the above diagram,
T + m₁a − m₁g = 0
T + 15a − 15g = 0
⇒ T = 15g − 15a                  (1)

Now,
T₁ − m₂g − m₂a = 0
T₁ − 5g − 5a = 0
⇒ T₁= 5g + 5a                                   (2)

Again,
 T − (T₁ + 5a + m₂R) = 0
⇒ T − (5g + 5a + 5a +m₂R) = 0        (3)
(where R = μg)

From Equations (1) and (2),
15g − 15a = 5g + 10a + 0.2 (5g)
⇒ 25a = 90          [g = 10 m/s²]
⇒ a = 3.6 m/s²

From Equation (3),
T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10 = 96 N in the left string.

From Equation (2),
T₁ = 5g + 5a
   = 5 × 10 + 5 × 36
   = 50 + 18
   = 68 N in the right string.

Question 15:

Given,
initial velocity of the vehicle, u = 36 km/h = 10 m/s
final velocity of the vehicle, v = 0
s = 5 m, $\mathrm{\mu }=\frac{4}{3}$, g = 10 m/s²

Let the maximum angle of incline be $\theta$.

Using the equation of motion
$$\begin{gathered} a=\frac{v^2-u^2}{2s}=\frac{0-{10}^2}{2\times 5} \\ =-10 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

From the free body diagram

R − mg cos θ = 0
⇒ R = mg cos θ                     (1)
Again,
ma + mg sin θ − μ R = 0
⇒ ma + mg sin θ − μmg cos θ = 0
⇒ a + g sin θ − μg cos θ = 0
$\Rightarrow 10+10 \sin \mathrm{\theta }-\left(\frac{4}{3}\right)\times 10 \cos \mathrm{\theta }=0$
⇒ 30 + 30 sin θ − 40 cos θ = 0
⇒ 3 + 3 sin θ − 4 cos θ = 0
⇒ 4 cos θ − 3 sin θ = 3
$\Rightarrow 4\sqrt{\left(1-{\sin }^2 \mathrm{\theta }\right)}=3+3 \sin \mathrm{\theta }$
On squaring, we get
16 (1 − sin² θ) = 9 + 9 sin² θ + 18 sin θ
25 sin² θ + 18 sin θ − 7 = 0
$$\begin{gathered} \Rightarrow \sin \mathrm{\theta }=\frac{18+\sqrt{{18}^2-4\left(25\right) \left(-7\right)}}{2\times 25} \\ =\frac{-18+32}{50}=\frac{14}{50}=0.28 \left( \mathrm{Taking} \mathrm{positve} \mathrm{sign} \mathrm{only}\right) \\ \Rightarrow \mathrm{\theta }={\sin }^{-1} \left(0.28\right)=16° \end{gathered}$$
Therefore, the maximum incline of the road, θ = 16°.

Question 16:

To reach the 50 m distance in minimum time, the superman has to move with maximum possible acceleration.
Suppose the maximum acceleration required is 'a'.
∴ ma − μR = 0 ⇒ ma = μ mg
⇒ a = μg = 0.9 × 10 = 9 m/s²


(a) As per the question, the initial velocity,
u = 0, t = ?
a = 9 m/s², s = 50 m

From the equation of motion,
$s=ut+\left(\frac{1}{2}\right)a{t}^2$
$$\begin{gathered} 50=0+\left(\frac{1}{2}\right)9t^2 \\ \Rightarrow t=\frac{10}{3} \mathrm{s} \end{gathered}$$

(b) After covering 50 m, the velocity of the athelete is
v = u + at
  $$\begin{gathered} =0+9\times \left(\frac{10}{3}\right) \mathrm{m}/\mathrm{s} \\ =30 \mathrm{m}/\mathrm{s} \end{gathered}$$
The superman has to stop in minimum time. So, the deceleration, a = − 9 m/s² (max)
R = mg
ma = μR                   (maximum frictional force)
ma = μmg 
⇒ a = μg
      = 9 m/s²  (deceleration)
u₁ = 30 m/s, v = 0
$$\begin{gathered} \Rightarrow t=\frac{v_1-u_1}{a} \\ =\frac{0-30}{-a} \\ =\frac{-30}{-a}=\frac{10}{3} \mathrm{s} \end{gathered}$$

Question 17:

When the driver applies hard brakes, it signifies that maximum force of friction is developed between the tyres of the car and the road.
So, maximum frictional force = μR

From the free body diagram,

R − mg cos θ = 0
⇒ R = mg cos θ                              (1)
and
μR + ma − mg sin θ = 0                   (2)
⇒ μ mg cos θ + ma − mg sin θ = 0


where θ = 30Ëš
$$\begin{gathered} \Rightarrow \mathrm{\mu }g \cos \mathrm{\theta }+a-10\times \left(\frac{1}{2}\right)=0 \\ \Rightarrow a=5-\left[\frac{1}{2}\sqrt{3}\right]\times 10\left(\frac{\sqrt{3}}{2}\right) \\ =5-\frac{10\times 3}{4} \\ =\frac{20}{4}-\frac{10\times 3}{4} \\ =\frac{-10}{4} \\ =-2.5 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
s = 12.8 m
u = 6 m/s
∴ Velocity at the end of incline
$$\begin{gathered} \nu =\sqrt{u^2+2as} \\ =\sqrt{6^2+2\left(2.5\right) \left(12.8\right)} \\ =\sqrt{36+64} \\ =10 \mathrm{m}/\mathrm{s}=36 \mathrm{km}/\mathrm{h} \end{gathered}$$
Therefore, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground, i.e. at 36 km/h.

Question 18:

Let a be the maximum acceleration of the car for crossing the bridge.



From the above diagram,
ma = μR
(For more accelerations the tyres will slip)
ma = μmg
a = μg = 1 × 10 = 10 m/s²

To cross the bridge in minimum possible time, the car must be at its maximum acceleration.
u = 0, s = 500 m, a = 10 m/s²

From the equation of motion,
$s=ut+\frac{1}{2}a{t}^2$
Substituting respective values
$$\begin{gathered} 500=0+\left(\frac{1}{2}\right)10t^2 \\ \Rightarrow t=\sqrt{100}=10 \mathrm{s} \end{gathered}$$
Therefore, if the car's acceleration is less than 10 m/s², it will take more than 10 s to cross the bridge. So, one cannot drive through the bridge in less than 10 s.

Question 19:

       
   
(a) From the free body diagram

       

R = 4g cos 30°
$$\begin{gathered} \Rightarrow \mathrm{R}=4\times 10\times \frac{\sqrt{3}}{2} \\ =20\sqrt{3} \mathrm{N} \left(1\right) \end{gathered}$$

μ₂R + m₁a − p − m₁g sin θ = 0
μ₂R + 4a − p − 4g sin 30° = 0              
⇒ 0.3 $\times$ (40) cos 30° + 4a − p − 40 sin 30° = 20             (2)

      

R₁ = 2g cos 30° $=10\sqrt{3}$                        (3)
p + 2a − μ₁R₁ − 2g sin 30° = 0               (4)

From Equation (2),
$6\sqrt{3}+4a-p-20=0$

From Equation (4),
$$\begin{gathered} p+2a+2\sqrt{3}-10=10 \\ 6\sqrt{3}+6a+30+2\sqrt{3}=0 \\ \Rightarrow 6a=30-8\sqrt{3} \\ =30-13.85=16.15 \\ \Rightarrow a=\frac{16.15}{6} \\ =2.69=2.7 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

(b) In this case, the 4 kg block will move at a higher acceleration because the coefficient of friction is less than that of the 2 kg block. Therefore, the two blocks will move separately. By drawing the free body diagram of 2 kg mass, it can be shown that a = 2.4 m/s².

Question 20:

     
From the free body diagram

   
R₁ = M₁g cos θ                                             (1)
R₂ = M₂g cos θ                                             (2)
T + M₁g sin θ − M₁a − μR₁ = 0                      (3)
T − M₂g + M₂a + μR₂ = 0                             (4)

From Equation (3),
T + M₁g sin θ − M₁a − μM₁g cos θ  = 0         (5)

From Equation (4),
T − M₂g sin θ + M₂a + μM₂ g cos θ = 0        (6)

From Equations (5) and (6),
g sin θ(M₁ + M₂) − a(M₁ + M₂) − μg cos θ(M₁ + M₂)
⇒ a(M₁ + M₂) = g sin θ(M₁ + M₂) = μg cos θ(M₁ + M₂)
⇒ a = g(sin θ − μ cos θ)a − g(sin θ − μ cos θ)
∴ The acceleration of the block (system) = g(sin θ − μcos θ)

The force exerted by the rod on one of the blocks is tension, T.
T = −M₁g sin θ + M₁a + μM₁g cos θ
T = −M₁g sin θ + M₁(g sin θ − μg cos θ) + μM₁g cos θ = 0

Question 21:

Let P be the force applied to slide the block at an angle θ.

        

From the free body diagram,
R + P sin θ − mg = 0
⇒ R = −P sin θ + mg            (1)
μR = P cos θ                        (2)

From Equation (1),
μ(mg − P sin θ)−P cos θ = 0
⇒ μmg = μP sin θ + P cos θ
$\Rightarrow P=\frac{\mathrm{\mu } mg}{\mathrm{\mu } \sin \mathrm{\theta }+\cos \mathrm{\theta }}$
The applied force P should be minimum, when μ sin θ + cos θ is maximum.
Again, μ sin θ + cos θ is maximum when its derivative is zero:
$\frac{d}{d\mathrm{\theta }}\left(\mathrm{\mu } \sin \mathrm{\theta }+\cos \mathrm{\theta }\right)=0$
⇒ μ cos θ − sin θ = 0
θ = tan⁻¹ μ
So, $P=\frac{\mathrm{\mu } mg}{\mathrm{\mu } \sin \mathrm{\theta }+\cos \mathrm{\theta }}$
Dividing numerator and denominator by cos θ, we get
       $=\frac{\mathrm{\mu } mg/\cos \mathrm{\theta }}{{\displaystyle \frac{\mathrm{\mu } \sin \mathrm{\theta }}{\cos \mathrm{\theta }}}+{\displaystyle \frac{\cos \mathrm{\theta }}{\cos \mathrm{\theta }}}}$

$$\begin{gathered} P=\frac{\mathrm{\mu } mg \sec \mathrm{\theta }}{1+\mathrm{\mu } \tan \mathrm{\theta }} \\ =\frac{\mathrm{\mu } mg \sec \mathrm{\theta }}{1+{\tan }^2 \mathrm{\theta }}=\frac{\mathrm{\mu } mg}{\sec \mathrm{\theta }} \\ =\frac{\mathrm{\mu } mg}{\sqrt{\left(1+{\tan }^2 \mathrm{\theta }\right)}}=\frac{\mathrm{\mu } mg}{\left(1+{\mathrm{\mu }}^2\right)} \end{gathered}$$(using the property $1+{\tan }^2\theta ={\sec }^2\theta$)
Therefore, the minimum force required is $\frac{\mathrm{\mu } mg}{\sqrt{\left(1+{\mathrm{\mu }}^2\right)}}$ at an angle θ = tan⁻¹ μ.

Question 22:

Let T be the maximum force exerted by the man on the rope.

From the free body diagram,
R + T = Mg
⇒ R = Mg − T                     (1)
Again,
R₁ − R − mg = 0
⇒ R₁ = R + mg                   (2)
and
T − μR₁ = 0

From Equation (2),
T − μ(R + mg) = 0
⇒ T − μR − μ mg = 0
⇒ T − μ(Mg − T) − μmg = 0
T − μMg + μt − μmg = 0



⇒ T (1 + μ) = μMg + μmg
$\Rightarrow T=\frac{\mathrm{\mu } \left(M+m\right)g}{1+\mathrm{\mu }}$
Therefore, the maximum force exerted by the man is $\frac{\mathrm{\mu }\left(M+m\right)g}{1+\mathrm{\mu }}$.

Question 23:

Consider the free body diagram.

(a) For the mass of 2 kg, we have:
R₁ − 2g = 0
⇒ R₁ = 2 × 10 = 20
2a + 0.2 R₁ − 12 = 0
⇒ 2a + 0.2 (20) = 12
⇒ 2a = 12 − 4
⇒ a = 4 m/s²

Now,
4a − μR₁ = 0
⇒ 4a = μR₁ = 0.2 (20) = 4
⇒ a₁ = 1 m/s²



The 2 kg block has acceleration 4 m/s² and the 4 kg block has acceleration 1 m/s².

(ii) We have:
R₁ = 2g = 20
Ma = μR₁ = 0
a = 0
And,
Ma + μmg − F = 0
4a + 0.2 × 2 × 10 − 12 = 0
⇒ 4a + 4 = 12
⇒ 4a = 8
⇒ a = 2 m/s²

Question 24:

Given:
μ₁ = 0.2
μ₂ = 0.3
μ₃ = 0.4
Using the free body diagram, we have:



(a) When the 10 N force is applied to the 2 kg block, it experiences maximum frictional force.
Here,
μ₁R₁ = μ₁ × m₁g
μ₁R₁ = μ₁ × 2g = (0.2) × 20
        = 4 N              (From the 3 kg block)
Net force experienced by the 2 kg block = 10 − 4 = 6 N
∴ $a_1=\frac{6}{2}=3 \mathrm{m}/{\mathrm{s}}^2$
But for the 3 kg block (Fig. 3), the frictional force from the 2 kg block, i.e, 4 N, becomes the driving force and the maximum frictional force between the 3 kg and 7 kg blocks.
Thus, we have:
μ₂ = R₂ = μ₂m₂g = (0.3) × 5 kg
     = 15 N

Therefore, the 3 kg block cannot move relative to the 7 kg block.
The 3 kg block and the 7 kg block have the same acceleration (a₂ = a₃), which is due to the 4 N force because there is no friction from the floor.
$\therefore a_2=a_3=\frac{4}{10}=0.4 \mathrm{m}/{\mathrm{s}}^2$

(b) When the 10 N force is applied to the 3 kg block, it experiences maximum frictional force of (15 + 4) N, i.e., 19 N, from the 2 kg block and the 7 kg block.
So, it cannot move with respect to them.
As the floor is frictionless, all the three bodies move together.
$\therefore a_1=a_2=a_3=\frac{10}{12}=\left(\frac{5}{6}\right) \mathrm{m}/{\mathrm{s}}^2$

(c) Similarly, it can be proved that when the 10 N force is applied to the 7 kg block, all three blocks move together with the same acceleration.
$\therefore a_1=a_2=a_3=\left(\frac{5}{6}\right) \mathrm{m}/{\mathrm{s}}^2$

Question 25:

From the free body diagrams of the two blocks, we have
R₁ = mg    ...(i)
F = μR₁+T    ...(ii)
T − μR₁ = 0    ..(iii)
From equations (i) and (ii), we have
F − μmg = T            ...(ii) 
From equations (i) and (iii), we have
T = μmg
Putting T = μmg in equation (ii), we have
F = μmg + μmg = 2μmg


(b) From the free body diagram of upper block, we have
2F − T − μmg = ma       ....(i)
From the free body diagram of lower block, we have
T = Ma + μmg

Putting the value of T in (i), we get
2F − Ma − μmg − μmg = ma
Putting F = 2μmg, we get
2(2μmg) − 2μmg = a(M + m)
⇒ 4μmg − 2μmg = a(M + m)
$\Rightarrow a=\frac{2 \mathrm{\mu }mg}{M+m}$  in opposite directions.

Question 26:

From the free body diagram, we have:
R₁ + ma − mg = 0
⇒ R₁ = m (g − a)
         = mg − ma                    ...(i)
Now,
F − T − μR₁ = 0 and
T − μR₁ = 0
⇒ F − [μ (mg − ma)] − μ(mg − ma) = 0
⇒ F − μ mg − μma − μmg + μma = 0
⇒ F = 2 μmg − 2 μma
       = 2 μm (g − a)

(b) Let the acceleration of the blocks be a₁.

R₁ = mg − ma               ....(i)
And,
2F −T − μR₁ = ma₁    ...(ii)
Now,
T = μR₁ + Ma₁
   = μmg − μma + Ma₁

Substituting the value of F and T in equation (ii), we get:
2[2μm(g − a)] − (μmg − μma + Ma₁) − μmg + μma = ma₁
⇒ 4μmg − 4μma − 2μmg + 2μma= ma₁ + Ma₁
$\Rightarrow a_1=\frac{2\mathrm{\mu }m \left(g-a\right)}{M+m}$
Thus, both the blocks move with same acceleration a₁ but in opposite directions.